A piece of wire 42 cm long is bent into the shape of a rectangle whose width is twice its length. Find the dimensions of the rectangle.
With out getting into the semantics of whether the width can be longer than the length - if we write the equation 2l=w (two lengths equal one width) and the equation 2l+2w=42 and substitute 2l from the first equation for w in the second equation we get this equation
2l+4l=42 now add like terms
6l=42 and then divide both sides of the equation by 6
l=7 so if the length is 7, then the width must be 14 (2*7) and we can answer the question
The width is 14 and the length is 7.
We check this by substituting 7 and 14 into the original second equation 2*7+2*14=42
do the arithmetic 14+28=42 => 42=42 it checks.
The length of the wire used to make the rectangle is 42 cm
That means that the circumference of the rectangle is 42
But the circumference of a rectangle = 2L + 2W where L is the length and W is the width
Then 2L +2W = 42
But we know that W = 2L
Now substitute w=2L
==> 2L + 2(2L )= 42
==> 6L = 42
==> L = 42/6= 7 cm
Then W = 2(7) = 14 cm
To check the answer:
2L + 2w = 2(7) + 2(14) = 14 + 28 = 42 cm